c# qr code generator Inside Microsoft SQL Server 2008: T-SQL Querying in C#

Creation QR Code JIS X 0510 in C# Inside Microsoft SQL Server 2008: T-SQL Querying

Inside Microsoft SQL Server 2008: T-SQL Querying
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HIERARCHYID values can become long when you keep adding new nodes between existing nodes whose canonical paths have consecutive last numbers. For example, say you have nodes with canonical paths /1/ and /2/ and you add a node between them. You get a new value whose canonical path is /1.1/. Now add a value between /1.1/ and /2/, and you get /1.2/. Now add a value between /1.1/ and /1.2/, and you get /1.1.1/. As you see, if you keep adding nodes between existing nodes in this manner, you can get lengthy paths (which represent lengthy HIERARCHYID values) even when the tree is not deep. If order among siblings is not important, you can always make sure to add new child nodes after the last existing child or before the rst one; this way, the paths are more economical. But when order among siblings matters, you can t control this. If you must frequently add new nodes between existing ones, you may end up with very long HIERARCHYID values. In such a case, you can periodically run a procedure, which I will provide here, that normalizes the HIERARCHYID values for the whole graph, making them shorter. Run the following code to create a new version of the AddEmp stored procedure:
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---------------------------------------------------------------------- Stored Procedure: AddEmp, -Inserts new employee who manages no one into the table --------------------------------------------------------------------IF OBJECT_ID('dbo.AddEmp', 'P') IS NOT NULL DROP PROC dbo.AddEmp; GO CREATE PROC dbo.AddEmp @empid AS INT, @mgrid AS INT, @leftempid AS INT, @rightempid AS INT, @empname AS VARCHAR(25) , @salary AS MONEY = 1000 AS DECLARE @hid AS HIERARCHYID; IF @mgrid IS NULL SET @hid = hierarchyid::GetRoot(); ELSE SET @hid = (SELECT hid FROM dbo.Employees WHERE empid = @mgrid).GetDescendant ( (SELECT hid FROM dbo.Employees WHERE empid = @leftempid), (SELECT hid FROM dbo.Employees WHERE empid = @rightempid) ); INSERT INTO dbo.Employees(empid, hid, empname, salary) VALUES(@empid, @hid, @empname, @salary); GO
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This version accepts the IDs of the two child employees between which you want to add the new one.
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Graphs, Trees, Hierarchies, and Recursive Queries
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Next, run the following code, which truncates the Employees table and populates it with data in such a manner that lengthy paths are produced:
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TRUNCATE TABLE dbo.Employees; EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = EXEC dbo.AddEmp @empname = @empid 'A'; @empid 'B'; @empid 'C'; @empid 'D'; @empid 'E'; @empid 'F'; @empid 'G'; @empid 'H'; @empid 'I'; @empid 'J'; @empid 'K'; @empid 'J'; @empid 'L'; @empid 'M'; @empid 'N'; @empid 'O'; @empid 'P'; @empid 'Q'; @empid 'E'; @empid 'S'; @empid 'T'; = = = = = = = = = 1, @mgrid = NULL, @leftempid = NULL, @rightempid = NULL, 2, @mgrid = 3, @mgrid = 4, @mgrid = 5, @mgrid = 6, @mgrid = 7, @mgrid = 8, @mgrid = 9, @mgrid = 1, @leftempid = NULL, @rightempid = NULL, 1, @leftempid = 1, @leftempid = 1, @leftempid = 1, @leftempid = 1, @leftempid = 1, @leftempid = 2, @rightempid = NULL, 2, @rightempid = 4, @rightempid = 4, @rightempid = 6, @rightempid = 6, @rightempid = 3, 3, 5, 5, 7,
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8, @leftempid = NULL, @rightempid = NULL, 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 8, @leftempid = 9, @rightempid = NULL, 9, @rightempid = 11, @rightempid = 11, @rightempid = 13, @rightempid = 13, @rightempid = 15, @rightempid = 15, @rightempid = 17, @rightempid = 17, @rightempid = 19, @rightempid = 19, @rightempid = 10, 10, 12, 12, 14, 14, 16, 16, 18, 18, 20,
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= 10, @mgrid = = 11, @mgrid = = 12, @mgrid = = 13, @mgrid = = 14, @mgrid = = 15, @mgrid = = 16, @mgrid = = 17, @mgrid = = 18, @mgrid = = 19, @mgrid = = 20, @mgrid = = 21, @mgrid =
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Then run the following code to show the current HIERARCHYID values and their canonical paths:
SELECT empid, REPLICATE(' | ', lvl) + empname AS emp, hid, hid.ToString() AS path FROM dbo.Employees ORDER BY hid;
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